MHTCET 16 May 2024 Shift 1 Paper

Concept:The scalar triple product of the vectors $(\bar{a} \times \bar{b})$, $(\bar{b} \times \bar{c})$, and $(\bar{c} \times \bar{a})$ equals $[\bar{a}\,\bar{b}\,\bar{c}]^2$.Explanation:Let $[\bar{x}\,\bar{y}\,\bar{z}] = \bar{x} \cdot (\bar{y} \times \bar{z})$ denote the scalar triple product.We need to evaluate $[\bar{a} \times \bar{b}\ \ \bar{b} \times \bar{c}\ \ \bar{c} \times \bar{a}]$.Write it as $(\bar{a} \times \bar{b}) \cdot \big((\bar{b} \times \bar{c}) \times (\bar{c} \times \bar{a})\big)$.Using the vector triple product identity $(\bar{u} \times \bar{v}) \times \bar{w} = ((\bar{u} \times \bar{v}) \cdot \bar{w})\,\bar{v} - ((\bar{u} \times \bar{v}) \cdot \bar{v})\,\bar{w}$, we get:$(\bar{b} \times \bar{c}) \times (\bar{c} \times \bar{a}) = \big((\bar{b} \times \bar{c}) \cdot \bar{a}\big)\,\bar{c} - \big((\bar{b} \times \bar{c}) \cdot \bar{c}\big)\,\bar{a}$.Since $(\bar{b} \times \bar{c}) \cdot \bar{c} = 0$, the second term vanishes.Thus the expression becomes $(\bar{a} \times \bar{b}) \cdot \big[\big((\bar{b} \times \bar{c}) \cdot \bar{a}\big)\bar{c}\big]$.This equals $\big((\bar{a} \times \bar{b}) \cdot \bar{c}\big)\,\big((\bar{b} \times \bar{c}) \cdot \bar{a}\big)$.Now $(\bar{a} \times \bar{b}) \cdot \bar{c} = [\bar{a}\,\bar{b}\,\bar{c}]$ and $(\bar{b} \times \bar{c}) \cdot \bar{a} = [\bar{b}\,\bar{c}\,\bar{a}] = [\bar{a}\,\bar{b}\,\bar{c}]$, using cyclic symmetry.Therefore the given triple product equals $[\bar{a}\,\bar{b}\,\bar{c}]^2$.Comparing with $\lambda [\bar{a}\,\bar{b}\,\bar{c}]^2$, we get $\lambda = 1$.Answer:$\lambda = 1$Hence, the correct option is C. 1.